Uniform convergence of characteristic functions implies uniform convergence of distribution

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Let $F(x)$ and $(F_{n})_{n\geq 1}$ be some distribution functions and let $\varphi(t)$ and $(\varphi_{n})_{n\geq 1}$ be their respective characteristic functions. I am trying to show that if:

$\sup_t |\varphi_n(t) -\varphi(t)| \rightarrow 0$, then

$\sup_x |F_n(x) -F(x)| \rightarrow 0$.

By Levy's theorem, I know that $F_n$ converges weakly to $F$. And I also know that, if $F$ is continuous, I have the result I want. So I believe I should use the uniform convergence of the characteristic functions to show that $F$ must be continuous.

I am trying to show that I must have:

$\int_{-\infty}^{\infty} |\varphi(t)|dt < \infty$

which would imply that $F$ is continuous, but I am stuck at that. Anyone can shed some light? Thanks!